The location of information about chemical elements and the use of that information for calculations or other purposes is a time-consuming as well as error prone process. Persons interested in working on complex chemical problems or calculations that are based on chemical element information are hindered by the time and convenience of locating the necessary information. Although there are available computer programs or calculators that assist in the retrieval and processing of this information, none are available in a convenient and portable form (such as found in a scientific calculator).
The following describes the types of problems that a chemical calculator could solve for the beginning chemistry student to a professional chemist, chemical engineer or other persons having a need to solve these kinds of chemical problems. The questions below are those that are typically found in chemistry textbooks or chemistry reference works. The steps required to answer these questions are followed by a brief notation as to the types of difficulties likely to be encountered or sources of error.
Question #1: What is the name, atomic number and mass, electron configuration and melting point of the element having the symbol "A1"? PA0 Answer: Aluminum, at.no.=13, at.mass=26.9815, e.configuration.=[Na]3s.sup.2 3p.sup.1, m.pt.=660.degree. C. PA0 Question #2: What is the difference in electronegativity and % ionic character in the bond between sodium and chlorine? PA0 Answer: E.N. diff.=3.0-1.0=2.0, % ionic character=63% PA0 Question #3: What is the constituent formula and formula weight of Al.sub.2 (SO.sub.4).sub.3 +2.5 H.sub.2 O? PA0 Answer: Formula=Al.sub.2 S.sub.3 O.sub.14.5, F.W.=387.18575 PA0 Question 4: What is the % (by weight) of the constituent elements found in the formula given in Question #3? PA0 Question #5: What is the a) (constituent) element formula and b) empirical formula corresponding to the following? EQU CH.sub.3 (CH.sub.2).sub.3 COCCl.sub.2 CO.sub.2 CH.sub.2 OH PA0 Answer: PA0 Question #6: What is the formula weight, constituent formula and % composition of the reactant A and product B shown in the following transformation? PA0 Question #7: How many grams of product B (Question #6) would be obtained from 5.16 g of A? PA0 Answer: 6.41 g of B PA0 Question #8: What is the empirical formula of the substance whose composition is: PA0 Answer: C.sub.4 H.sub.10 O.sub.3 S.sub.4 PA0 Question #9: What amount of product D is formed from 3.16 g of reactant C in the following chemical reaction? ##EQU5## Answer: 3.32 g (0.0313 moles) of D
Steps Involved in Answering This Question: Typically a person would check a periodic table or reference table (found in a reference handbook, separate attachment or wall chart). Any of these reference sources may not be readily available if the person is in a laboratory or plant, classroom or in the field. In addition, the reference source may not have all of the information, e.g. many condensed periodic tables may have only information about the first three. Even a large wall chart may not be readable to more than a few entries to a person at the back of a classroom.
Steps Involved in Answering This Question: Along with the problems of locating a source containing the electronegativity values (as described for Question #1 above), there is also the problem of knowing how to calculate the % ionic character.
Step 1: Write the formula
Step 2: Locate atomic mass of constituent elements
Step 3: Set up for calculations
______________________________________ 3.1 2 Al = 2 .times. 26.9813 = 53.9626 = 53.9626 3.2 1 S = 1 .times. 32.064 = 32.064 3.3 4 O = 4 .times. 15.9994 = 63.9976 3.4 3 (SO.sub.4) = 3 .times. 96.0616 = 288.1848 3.5 343.1474 3.6 2 H = 2 .times. 1.00797 = 2.01594 3.7 1 O = 1 .times. 15.9994 = 15.9994 3.8 2.5 (H.sub.2 O) = 2.5 .times. 18.01534 = 45.03835 3.9 Formula Weight = 387.18575 ______________________________________
Sources of Errors and Difficulties: There are some dozen separate steps or operations that must be performed to answer the question. Examples of the major types of errors are:
Steps 1-3: Incorrect formula entry and set-up so that subscripts and/or coefficients multiply proper quantities. In steps 3.3 and 3.4, for example, the atomic mass of oxygen must be multiplied (step 2.3) by 4 before the sum within the parentheses is multiplied by 3 (step 3.4).
Step 2: Involves the location and transcription of the correct atomic weight of the element (beginning students often use the atomic number instead of the atomic mass).
Step 3: The use of computers or math calculators does speed up the actual calculation process, but a common source of error arises from typographical errors in number entry. (For example, this author obtained two slightly different answers to the question the two times they were tried due to minor typographical errors.)
Step 3.8: The student must recognize the significance and operation of the coefficient "2.5" that is multiplied times the formula weight of H.sub.2 O and this added to the formula weight of the first part of the formula.
______________________________________ Answer: 13.937% Al 24.843% S 88.324% Al.sub.2 (SO.sub.4).sub.3 49.587% O plus 1.360% H 11.683% H.sub.2 O 10.331% O ______________________________________
Description of Process for Solution: The first steps in the calculations are similar to that used for Question #3, except that it is necessary to separate the weights constituted by each element from the remainder of the formula:
______________________________________ In the First Part of the Formula: 4.01: for Al, there are 2 Al = 2 .times. 26.9626 = 53.9626 4.02: for S, there are 3 S (1 S in each SO.sub.4) = 3 .times. 32.064 = 96.192 4.03: for O, there are 12 O (4 O in each SO.sub.4) = 12 .times. 15.9994 = 191.9928 In the Second Part of the Formula: 4.04: for H, there are 5 H (2 H in H.sub.2 O) = 2 .times. 2.5 .times. 1.00797 = 5.26485 4.05: for O, there are 2.5 O (1 O in H.sub.2 O) = 2.5 .times. 15.9994 = 39.9985 Formula Weight = 387.4108 ______________________________________
The % composition for each element is obtained by dividing the weights obtained in Steps 4.01-4.05 by the formula weight: EQU 4.06: ( 53.9626/387.4108).times.100=13.929% Al EQU 4.07: ( 96.192/387.4108).times.100=24.829% S EQU 4.08: (191.9928/387.4108).times.100=49.558% O EQU 4.09: ( 5.26485/387.4108).times.100=1.359% H EQU 4.10: ( 39.9985/387.4108).times.100=10.324 % O
Constituent element formula: C.sub.8 H.sub.12 O.sub.4 Cl.sub.2 PA1 Empirical formula: C.sub.4 H.sub.6 O.sub.2 Cl PA1 1. The constituent formula of reactant A is calculated following the procedures illustrated for the solution to Questions #4 and #5. The formula weight and % composition are then calculated using this formula as shown in the solution to Question #4. PA1 2. Typically, the person trying to determine the constituent formula and/or formula weight of product B would recognize that formula B is related to formula A by the "loss" of H.sub.2 O and "gain" of CH.sub.3 C.sub.2 OH[C.sub.2 H.sub.6 O] in the transformation and would set up the calculation as follows: ##STR5## The formula weight of B can then be determined by either subtracting and adding (respectively) the formula weights of H.sub.2 O (18.0152) and C.sub.2 H.sub.6 O (46.0688) from the formula weight of A: ##STR6## or by using the formula of B to calculate its formula weight: ##EQU1## PA1 % C=28.22 PA1 % H=5.92 PA1 % O=28.19 PA1 % S=37.66 PA1 A) the calculated formula weight corresponding to the chemical formula. PA1 B) the % composition (of the constituent elements) PA1 C) the chemical formula (of constituent elements).
Description of Process for Solution: To answer 5a), the procedure is relatively straight-forward, i.e. you simply count the number of atoms of each element: CH.sub.3 (CH.sub.2).sub.3 COCCl.sub.2 CO.sub.2 CH.sub.2 OH
__________________________________________________________________________ Counting: CH.sub.3 --(CH.sub.2).sub.3 --C--O--C--Cl.sub.2 --C--O.sub.2 --CH.sub.2 --O--H Total __________________________________________________________________________ for C: 131--1--1--1---- 8 C for H: 36------------2--1 12 H for O: ------1------2--1-- 4 O for Cl: ----------2---------- 2 Cl Formula = C.sub.8 H.sub.12 O.sub.4 Cl.sub.2 __________________________________________________________________________
Although this counting would appear to be rather straight-forward, the miscounting of only one element or summing error would produce a completely erroneous result.
To obtain an empirical formula, the constituent element formula obtained above must be examined to see if all of the subscripts are divisible by any number to produce the "simplest" empirical formula, i.e. that represent the formula that shows a ratio of the fewest atoms of the constituent elements that would be obtained by %-composition analysis (see Question #8 below). In this case, the subscripts are all divisible by 2 to give: C.sub.4 H.sub.6 O.sub.2 Cl.
______________________________________ ##STR1## ##STR2## ##STR3## ______________________________________ ##STR4## ______________________________________
Solution to Question #6:
While either of these procedures yield the same result, only the latter is useful in calculating the % composition: ##EQU2##
Suffice it to say, a number of steps must be taken and the choice of steps depends on the type of information or result desired. (The writer of this example spend in excess of 20 minutes setting up the mathematical operations to solve these questions. While the use of a calculator speeds up the process, it is not uncommon to still have a result that contains a mathematical error.)
Solution to Question #7: Typically, the solution to this problem is obtained by determining the number of moles of A corresponding to 5.16 g. ##EQU3## According to the transformation, 0.04442 moles of B should be formed, therefore: ##EQU4##
As before, two separate mathematical operations must be performed in order to answer this question. Additional calculations would be required if the coefficients of the reactants and products in the chemical reaction shown in Question #6 were not all the same.
Description of Process: The first step in answering this question involves dividing the % values for each element by the atomic weight of that element. This gives a formula in which the subscripts of the formula are proportional to the ratios of the atoms of the constituent elements in the compound: EQU 8.01 for C=28.22/12.01115=2.34948 EQU 8.02 for H=5.92/1.00797=5.87319 EQU 8.03 for O=28.19/15.9994=1.76194 EQU 8.04 for S=37.66/32.064=1.17452
First-decision Empirical Formula: EQU C.sub.2.34948 H.sub.5.87639 O.sub.1.76194 S.sub.1.117452
At this point, the goal is to reduce the formula to obtain whole number subscripts (i.e. you can not have "fractions" of atoms). While in come cases it might be possible to "round off" the numbers, typically this is not done at this point. Rather, the numbers are divided by the lowest value; in this case, 1.17452, for sulfur. EQU 8.05 for C=2.34948/1.17452=2.00037 EQU 8.06 for H=5.87369/1.17452=5.00050 EQU 8.07 for O=1.76194/1.17452=1.50014 EQU 8.08 for S=1.17452/1.17452=1.00000
The new formula, rounding off to 2 decimal places is now: EQU C.sub.2.00 H.sub.5.00 O.sub.1.50 S.sub.1.00
At this point, it seems obvious that to obtain simple whole-number ratios of atoms, the formula subscripts should be multiplied by 2 to give the empirical formula: EQU C.sub.4 H.sub.10 O.sub.2 S.sub.2
Suffice it to say, this process involves several points of decision as to division or multiplication operations or rounding-off. The repetition of these decisions and mathematical operations account for significant introduction of transmission errors.
Description of Process: The first step involves determining the number of moles of reactant C: ##EQU6## The formula weight of C must be determined as before. The second step involves setting up a ratio of moles C used to moles D produced according to the chemical equation: ##EQU7##
Again, the set-up required to answer this question is time-consuming and error-prone. Typically, the answer will be in error by some ratio of the coefficients of the reactants and products involved in the calculation.